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  • Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (iii) textbook solution.

Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (iii) textbook solution.

Answers (1)

Answer : y+2 x=3 x^{2} y

Hint               : using variable separable method and substituting the values

Given :   x^{2} d y+\left(x y+y^{2}\right) d x=0, \quad y=1, \text { when } x=1

Solution : x^{2} d y+\left(x y+y^{2}\right) d x=0

The differential equation can be  written as

\begin{aligned} &x^{2} d y=-\left(x y+y^{2}\right) d x\\ &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}                                      ......(i)

Let f(x, y)=\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}}

Finding  f(\lambda \; x,\lambda \; y)

\begin{aligned} f(\lambda x, \lambda y) &=\frac{-\left(\lambda x . \lambda y+\lambda^{2} y^{2}\right)}{\lambda^{2} x^{2}}=\frac{-\lambda^{2}\left(x y+y^{2}\right)}{\lambda^{2} x^{2}} \\ &=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &=\lambda^{0}(f(x, y)) \\ f(x, y)=& \frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}

Therefore , f(x,y) is a homogenous function of degree zero.

Putting y=vx

Diff w.r.t.x

\frac{d y}{d x}=x \frac{d v}{d x}+v

Putting value of \frac{d y}{d x} \text { and } y=v x \text { in }(i)

\begin{aligned} &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x(v x)+(v x)^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x^{2} v+x^{2} v^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-x^{2}\left(v+v^{2}\right)}{x^{2}} \end{aligned}

\begin{aligned} &x \frac{d v}{d x}=-v-v^{2}-v=-\left(v^{2}+2 v\right) \\ &\frac{d v}{v^{2}+2 v}=-\frac{d x}{x} \end{aligned}

Integrating both sides we get ,

\begin{aligned} &\int \frac{d v}{v^{2}+2 v}=-\int \frac{d x}{x} \\ \end{aligned}

\begin{aligned} &\int \frac{d v}{\left(v^{2}+2 v+1\right)-1}=-\log x+\log c \\ \end{aligned}

\begin{aligned} &\int \frac{d v}{(v+1)^{2}-1^{2}}=-\log x+\log c \\ \end{aligned}

Using \begin{aligned} & \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c \\ \end{aligned}

\begin{aligned} &\frac{1}{2} \log \frac{v+1-1}{v+1+1}=-\log x+\log c \end{aligned}

\begin{aligned} &\frac{1}{2} \log \frac{v}{v+2}=-\log x+\log c \end{aligned}

\begin{aligned} &\log \sqrt{\frac{v}{v+2}}+\log x=\log c \\ &\frac{x \sqrt{v}}{\sqrt{v+2}}=c \end{aligned}

Now putting value of v i.e , y/x

\begin{aligned} &\Rightarrow \frac{x \sqrt{\frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}

\begin{aligned} &\frac{\sqrt{x^{2} \times \frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}

\begin{aligned} &\frac{\sqrt{x y}}{\sqrt{\frac{y+2 x}{x}}}=c \\ \end{aligned}

\begin{aligned} &\frac{x \sqrt{y}}{\sqrt{y+2 x}}=c \\ \end{aligned}

\begin{aligned} &x \sqrt{y}=c \sqrt{y+2 x} \end{aligned}

Squaring both sides

\begin{aligned} &x^{2} y=c^{2}(y+2 x) \quad \; \; \; \; \; \; \; \; \; \ldots (ii)\\ &\text { Now } x=1, y=1\\ &\text { Therefore }\left(1^{2}\right)(1)=c^{2}(1+2(1)) \end{aligned}

\begin{aligned} &\Rightarrow 1=c^{2}(3) \\ &\Rightarrow c^{2}=\frac{1}{3} \end{aligned}

Putting back in (ii)

\begin{aligned} &\Rightarrow x^{2} y=\frac{1}{3}(y+2 x) \\ &3 x^{2} y=y+2 x \\ &y+2 x=3 x^{2} y \end{aligned}

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